Find $\lim_{x\to 0}\dfrac{x\cos(x)}{(x+1)\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac12$ (Choice B) B $1$ (Choice C) C $0$ (Choice D) D The limit doesn't exist.
Substituting $x=0$ into $\dfrac{x\cos(x)}{(x+1)\sin(x)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 0}\dfrac{x\cos(x)}{(x+1)\sin(x)} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[x\cos(x)]}{\dfrac{d}{dx}[(x+1)\sin(x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{-x\sin(x)+\cos(x)}{\sin(x)+(x+1)\cos(x)} \\\\ &=\dfrac{-(0)\sin(0)+\cos(0)}{\sin(0)+(0+1)\cos(0)} \gray{\text{Substitution}} \\\\ &=1 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[x\cos(x)]}{\dfrac{d}{dx}[(x+1)\sin(x)]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{x\cos(x)}{(x+1)\sin(x)}=1$.